Question 12.12: An outward flow reaction turbine running at 275 rpm under a ......
An outward flow reaction turbine running at 275 rpm under a net head of 200 m has internal and external runner diameters of 1.75 m and 2.5 m respectively. The water flow rate through the runner is 6.8 m^3/s. The constant width of the runner is 0.3 m. If the discharge at outlet is radial, find the flow velocities and vane angles at inlet and outlet of the runner vanes.
Data. Refer Figure 12.21. N=275 rmm ; H=200 m ; D_1=1.75 m ; D_2=2.5 m;Q=6.8 m ^3 / s ; B_1=B_2=0.3 m ; \beta=90^{\circ}
The tangential velocity of runner at inlet and outlet are:
u_1=\frac{\pi D_1 N}{60}=\frac{\pi \times 1.75 \times 275}{60}=25.2 m / s
u_2=\frac{\pi D_2 N}{60}=\frac{\pi \times 2.5 \times 275}{60}=34 m / s
From the equation, Q=\pi D_1 B_1 V_{f 1}, the flow velocities at inlet and outlet are:
\begin{gathered}V_{f 1}=\frac{Q}{\pi D_1 B_1}=\frac{6.8}{\pi \times 1.75 \times 0.3}=4.123 m / s \\V_{f 2}=\frac{Q}{\pi D_2 B_2}=\frac{6.8}{\pi \times 2.5 \times 0.3}=2.89 m / s\end{gathered}
The total head H can be written as the sum of the work done and kinetic energy at exit:
\begin{aligned}H &=\frac{V_{w 1}u_1}{g}+\frac{V_2^2}{2 g}=\frac{V_{w 1}u_1}{g}+\frac{V_{f 2}^2}{2 g}\quad\left(V_2=V_{f2}\right) \\200 &=\frac{V_{w 1}\times 25.2}{9.81}+\frac{2.89^2}{2 \times 9.81}\end{aligned}
Solving, whirl velocity at inlet, V_{w 1}=77.7 m / s
From the inlet velocity triangle,
\begin{aligned}\tan \theta &=\frac{V_{f 1}}{V_{w 1}-u_1}=\frac{4.123}{77.7-25.2}=0.0785 \\∴ \theta &=4.5^{\circ}\end{aligned}
From the outlet velocity triangle,
\begin{aligned}\tan \phi &=\frac{V_{f 2}}{u_2}=\frac{2.89}{34}=0.085 \\∴ \phi &=4.86^{\circ}\end{aligned}.
